【网络流+最短路】 HDU 5294 Tricks Device

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最短路求出到每个点的最小值以及边数

枚举下能到达终点的点的边数num[i] 取个最小值 +1  然后减去m就是第二个输出的数

对于原来的图满足最短路的边建上一条边跑网络流 就得到最小割= =

因为有重边所以网络流的边的容量为该边最小权值的数量

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <string>
#include <iostream>
#include <algorithm>
using namespace std;
#include <vector>
typedef long long LL;
using namespace std;
const LL mod=1e9+7;
const int MAXN = 2010;//点数的最大值
const int MAXM = 160000;//边数的最大值
const int INF = 0x3f3f3f3f;
struct Sap
{
struct Edge
{
int to,next,cap,flow;
} edge[MAXM]; //注意是MAXM
int tol;
int head[MAXN];
int gap[MAXN],dep[MAXN],pre[MAXN],cur[MAXN];
void init()
{
tol = 0;
memset(head,-1,sizeof (head));
}
//加边，单向图三个参数，双向图四个参数
void addedge (int u,int v,int w,int rw=0)
{
edge[tol].to = v;
edge[tol].cap = w;
edge[tol].next = head[u];
edge[tol].flow = 0;
head[u] = tol++;
edge[tol].to = u;
edge[tol].cap = rw;
edge[tol]. next = head[v];
edge[tol].flow = 0;
head[v]=tol++;
}
//输入参数：起点、终点、点的总数
//点的编号没有影响，只要输入点的总数
int sap(int start,int end, int N)
{
memset(gap,0,sizeof(gap));
memset(dep,0,sizeof(dep));
memcpy(cur,head,sizeof(head));
int u = start;
pre[u] = -1;
gap[0] = N;
int ans = 0;
int i;
while(dep[start] < N)
{
if(u == end)
{
int Min = INF;
for( i = pre[u]; i != -1; i = pre[edge[i^1]. to])
{
if(Min > edge[i].cap - edge[i]. flow)
Min = edge[i].cap - edge[i].flow;
}
for( i = pre[u]; i != -1; i = pre[edge[i^1]. to])
{
edge[i].flow += Min;
edge[i^1].flow -= Min;
}
u = start;
ans += Min;
continue;
}
bool flag = false;
int v;
for( i = cur[u]; i != -1; i = edge[i].next)
{
v = edge[i]. to;
if(edge[i].cap - edge[i].flow && dep[v]+1 == dep[u])
{
flag = true;
cur[u] = pre[v] = i;
break;
}
}
if(flag)
{
u = v;
continue;
}
int Min = N;
for( i = head[u]; i != -1; i = edge[i]. next)
{
if(edge[i].cap - edge[i].flow && dep[edge[i].to] < Min)
{
Min = dep[edge[i].to];
cur[u] = i;
}
}
gap[dep[u]]--;
if(!gap[dep[u]]) return ans;
dep[u] = Min+1;
gap[dep[u]]++;
if(u != start) u = edge[pre[u]^1].to;
}
return ans;
}
}wll;
bool vis[MAXN];
int pre[MAXN];
int cost[MAXN][MAXN],lowcost[MAXN],num[MAXN];
void Dijkstra(int n,int beg)
{
for(int i=0; i<n; i++)
{
num[i]=0;
lowcost[i]=INF;
vis[i]=false;
pre[i]=-1;
}
lowcost[beg]=0;
for(int j=0; j<n; j++)
{
int k=-1;
int Min=INF;
for(int i=0; i<n; i++)
if(!vis[i]&&lowcost[i]<Min)
{
Min=lowcost[i];
k=i;
}
if(k==-1)break;
vis[k]=true;
for(int i=0; i<n; i++)
{
if(!vis[i]&&lowcost[k]+cost[k][i]<lowcost[i])
{
num[i]=num[k]+1;
lowcost[i]=lowcost[k]+cost[k][i];
pre[i]=k;
}
else if(i!=k&&lowcost[k]+cost[k][i]==lowcost[i]&&num[i]>num[k])
{
pre[i]=k;
}
}
}
}
int val[MAXN][MAXN];
int main()
{
int n,m,a,b,c;
while(scanf("%d%d",&n,&m)!=EOF)
{
for(int i=0;i<n;i++)
{
for(int j=0;j<n;j++)
cost[i][j]=INF;
cost[i][i]=0;
}
for(int i=0;i<m;i++)
{
scanf("%d%d%d",&a,&b,&c);
a--,b--;
if(cost[a][b]>c)
{
cost[a][b]=cost[b][a]=c;
val[a][b]=val[b][a]=1;
}
else if(cost[a][b]==c)
val[a][b]++,val[b][a]++;
}
Dijkstra(n,0);
int ans=INF;
for(int i=0;i<n;i++)
if(lowcost[n-1]==lowcost[i]+cost[i][n-1]&&num[i]+1<ans)
ans=num[i]+1;
wll.init();
for(int i=0;i<n;i++)
for(int j=0;j<n;j++)
{
if(i==j) continue;
if(cost[i][j]+lowcost[i]==lowcost[j])
wll.addedge(i,j,val[i][j]);
}
int first1=wll.sap(0,n-1,n);
printf("%d %d\n",first1,m-ans);
}
return 0;
}
/*
3 5
1 2 1
1 2 1
1 2 2
2 3 1
2 3 1
*/